(18.2.1) The Fabry equation for solenoids
A solenoid has an inner radius $R_{1}$, an outer radius $R_{2}$, and a length $2 L$. The current is $I$.
(a) Show that at the center
$$
B=\mu_{0} n I L \ln \frac{\alpha+\left(\alpha^{2}+\beta^{2}\right)^{1 / 2}}{1+\left(1+\beta^{2}\right)^{1 / 2}}
$$
where $n$ is the number of turns per square meter $(\infty 1 /$ cross section of the wire), $\alpha=R_{2} / R_{1}$, and $\beta=L / R_{1}$.
(b) Show that the length of the wire is
$$
l=n V=2 \pi n\left(\alpha^{2}-1\right) \beta R_{1}^{3}
$$
where $V$ is the volume of the winding.
(c) Check the Fabry equation, which states that at the center of any solenoid
$$
B=G\left(\frac{P \lambda \sigma}{R_{1}}\right)^{1 / 2}
$$
Here $G$ depends on the geometry, $P$ is the dissipated power, $\lambda=n \pi r^{2}$ is the filing factor, or the fraction of the coil cross section occupied by the conductor, $r$ is the radius of the wire, and $o$ the resistivity.